Explanation Given: The outer radius of the shell is r 2 . The heat flux is q ˙ 1 . The heat transfer coefficient of h . The inner radius of the of shell is r 1 . The thermal conductivity of shell material is k . The air temperature is T ∞ . Concept used: Write the expression for heat conduction 3-Dimension equation for shell. 1 r 2 ∂ ∂ r ( r 2 ∂ T ∂ r ) + 1 r 2 ∂ ∂ ϕ ( ∂ T ∂ ϕ ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ T ∂ θ ) + e ˙ g k = 1 α δ T δ t ...... (1) Here, k is thermal conductivity, e ˙ g is the heat generation, and α is thermal diffusivity. Calculation: Here, the heat transfer is steady. 1 α δ T δ t = 0 The heat generation is zero. e ˙ g k = 0 k = 0 The heat transfer is one dimensional. 1 r 2 ∂ ∂ ϕ ( ∂ T ∂ ϕ ) = 0 and 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ T ∂ θ ) = 0 Therefore substitute 0 for e ˙ g k , 0 for 1 r 2 ∂ ∂ ϕ ( ∂ T ∂ ϕ ) , 0 for 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ T ∂ θ ) and 0 for 1 α δ T δ t in equation (1) d d r ( r 2 d T d r ) = 0 ...... (2) The boundary condition for this problem is, q ˙ 1 = − k d T ( r 1 ) d r ...... (3) Here, q ˙ 1 is the heat flux. The second boundary condition is, − k d T ( r 2 ) d r = h ( T 2 − T ∞ ) ...... (4) Here, T 2 is the inner surface temperature of shell and h is the heat transfer coefficient. Integrate the equation (2). ∫ d d r ( r 2 d T d r ) The expression for constant C 1 is , d T d r = C 1 r 2 Integrate the equation again, ∫ d T d r = C 1 r 2 The relation obtain for the variation of temperature is, T ( r ) = − C 1 r + C 2 .....

Heat and Mass Transfer: Fundamentals and Applications

5th Edition

ISBN: 9780073398181

Author: Yunus A. Cengel Dr., Afshin J. Ghajar

Publisher: McGraw-Hill Education

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Question

Chapter 2, Problem 73P

To determine

The variation of temperature in the shell wall and the outer surface temperature of shell.

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Chapter 2, Problem 72P

Chapter 2, Problem 74P

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Heat and Mass Transfer: Fundamentals and Applications

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